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Concept 08 / 10Sets, Relations and Functions

Special Functions: Modulus, Greatest Integer, Fractional Part and Signum

Definitions, graphs and working identities for |x|, [x], {x} and sgn x, the triangle inequality and its equality case, the distance meaning of |x - a|, and solving equations and inequalities that mix these functions.

Figure 1
These four piecewise functions appear throughout JEE, from domain and range questions to limits, continuity and integration. Each is simple to define, but each has properties that are easy to misuse. Modulus (absolute value): ∣x∣=x|x| = x if x≥0x \ge 0 and ∣x∣=−x|x| = -x if x<0x < 0. Equivalently ∣x∣=x2=max⁡(x,−x)|x| = \sqrt{x^{2}} = \max(x, -x). The best way to think about it is geometrically: ∣x−a∣|x - a| is the distance between xx and aa on the number line. That view makes inequalities obvious: ∣x−a∣<r  ⟺  a−r<x<a+r|x - a| < r \iff a - r < x < a + r, and ∣x−a∣>r  ⟺  x<a−r|x - a| > r \iff x < a - r or x>a+rx > a + r. Algebraic facts: ∣xy∣=∣x∣∣y∣|xy| = |x||y|, ∣xy∣=∣x∣∣y∣|\frac{x}{y}| = \frac{|x|}{|y|}, and ∣x∣2=x2|x|^{2} = x^{2}. The triangle inequality states ∣a+b∣≤∣a∣+∣b∣|a + b| \le |a| + |b|, with equality if and only if ab≥0ab \ge 0 (the numbers do not point in opposite directions). Its companion is ∣a−b∣≥∣∣a∣−∣b∣∣|a - b| \ge ||a| - |b||. The graph of y=∣x∣y = |x| is a V with its vertex at the origin, and y=∣f(x)∣y = |f(x)| is obtained from y=f(x)y = f(x) by reflecting in the xx-axis every part that lies below it. Greatest integer (floor) function: [x][x], also written ⌊x⌋\lfloor x \rfloor, is the greatest integer less than or equal to xx. So [2.7]=2[2.7] = 2, [5]=5[5] = 5 and, importantly, [−2.3]=−3[-2.3] = -3, not −2-2, since −3-3 is the greatest integer that does not exceed −2.3-2.3. Its defining property is the sandwich [x]≤x<[x]+1[x] \le x < [x] + 1, or equivalently x−1<[x]≤xx - 1 < [x] \le x. The graph is a staircase: constant on each [n,n+1)[n, n + 1) and jumping up by 11 at every integer. At each step the left end is filled and the right end is open. Identities: [x+n]=[x]+n[x + n] = [x] + n for any integer nn (you may pull integers out); [x]+[−x]=0[x] + [-x] = 0 if xx is an integer and −1-1 otherwise; [x+y]∈{[x]+[y],[x]+[y]+1}[x + y] \in \{[x] + [y], [x] + [y] + 1\}; and Hermite's identity [x]+[x+12]=[2x][x] + [x + \frac{1}{2}] = [2x]. To solve an equation involving [x][x], set [x]=n[x] = n, an integer, solve for nn, and convert back using n≤x<n+1n \le x < n + 1. Fractional part: {x}=x−[x]\{x\} = x - [x], so x=[x]+{x}x = [x] + \{x\} always, and 0≤{x}<10 \le \{x\} < 1. For example {2.7}=0.7\{2.7\} = 0.7 and {−2.3}=−2.3−(−3)=0.7\{-2.3\} = -2.3 - (-3) = 0.7. Note that the fractional part of a negative number is not simply the digits after the decimal point. The graph is a sawtooth: on each [n,n+1)[n, n + 1) it is the line y=x−ny = x - n, rising from 00 towards 11 and then dropping back. It is periodic with period 11. Also {x}+{−x}=0\{x\} + \{-x\} = 0 if xx is an integer and 11 otherwise. Signum: sgn(x)=1\text{sgn}(x) = 1 for x>0x > 0, 00 for x=0x = 0, and −1-1 for x<0x < 0. For x≠0x \neq 0, sgn(x)=∣x∣x=x∣x∣\text{sgn}(x) = \frac{|x|}{x} = \frac{x}{|x|}, and in general x=∣x∣⋅sgn(x)x = |x| \cdot \text{sgn}(x). Composites such as sgn(x2−4)\text{sgn}(x^{2} - 4) are best handled by drawing a sign chart for the inner expression. How these interact with other ideas. [x][x] and {x}\{x\} are discontinuous at every integer, so they appear constantly in limit and continuity questions: at x=nx = n, [x][x] has left limit n−1n - 1 and right limit nn. ∣x∣|x| is continuous everywhere but not differentiable at 00. In integrals, the standard move is to split the interval at the integers, where [x][x] is constant on each piece. Key Takeaways: - ∣x−a∣|x - a| is distance; ∣x−a∣<r  ⟺  a−r<x<a+r|x - a| < r \iff a - r < x < a + r. - ∣a+b∣≤∣a∣+∣b∣|a + b| \le |a| + |b|, with equality iff ab≥0ab \ge 0; also ∣a−b∣≥∣∣a∣−∣b∣∣|a - b| \ge ||a| - |b||. - [x]≤x<[x]+1[x] \le x < [x] + 1; [x+n]=[x]+n[x + n] = [x] + n for integer nn; [−2.3]=−3[-2.3] = -3. - [x]+[−x]=0[x] + [-x] = 0 for integer xx and −1-1 otherwise; {x}+{−x}=0\{x\} + \{-x\} = 0 or 11 correspondingly. - {x}=x−[x]∈[0,1)\{x\} = x - [x] \in [0, 1); {x}\{x\} has period 11. - sgn(x)=∣x∣x\text{sgn}(x) = \frac{|x|}{x} for x≠0x \neq 0, and sgn(0)=0\text{sgn}(0) = 0. - To solve with [x][x]: let [x]=n∈Z[x] = n \in \mathbb{Z}, then use n≤x<n+1n \le x < n + 1. - Pitfall: [x]⋅[y]≠[xy][x] \cdot [y] \neq [xy] and [2x]≠2[x][2x] \neq 2[x] in general (take x=0.5x = 0.5). Brain-Teaser 1: Solve [x]2−5[x]+6=0[x]^{2} - 5[x] + 6 = 0, and then solve [x2]−5[x]+6=0[x^{2}] - 5[x] + 6 = 0 for x≥0x \ge 0. Solution: In the first equation let n=[x]n = [x]. Then n2−5n+6=0n^{2} - 5n + 6 = 0 gives n=2n = 2 or n=3n = 3. Now [x]=2  ⟺  2≤x<3[x] = 2 \iff 2 \le x < 3 and [x]=3  ⟺  3≤x<4[x] = 3 \iff 3 \le x < 4, so the solution set is the whole interval [2,4)[2, 4). An equation in [x][x] usually has intervals of solutions, not isolated points. The second equation is different, because [x2]≠[x]2[x^{2}] \neq [x]^{2}. Let [x]=n[x] = n with n≤x<n+1n \le x < n + 1; then n2≤x2<(n+1)2n^{2} \le x^{2} < (n + 1)^{2}, so [x2][x^{2}] lies between n2n^{2} and n2+2nn^{2} + 2n. The equation requires [x2]=5n−6[x^{2}] = 5n - 6, so n2≤5n−6≤n2+2nn^{2} \le 5n - 6 \le n^{2} + 2n. The left part gives 2≤n≤32 \le n \le 3, and the right part n2−3n+6≥0n^{2} - 3n + 6 \ge 0 always holds. For n=2n = 2: [x2]=4[x^{2}] = 4 with x∈[2,3)x \in [2, 3), so x2∈[4,5)x^{2} \in [4, 5), giving x∈[2,5)x \in [2, \sqrt{5}). For n=3n = 3: [x2]=9[x^{2}] = 9 with x∈[3,4)x \in [3, 4), so x∈[3,10)x \in [3, \sqrt{10}). So the solution set is [2,5)∪[3,10)[2, \sqrt{5}) \cup [3, \sqrt{10}). Brain-Teaser 2: Solve 2[x]=x+{x}2[x] = x + \{x\}. Solution: Substitute x=[x]+{x}x = [x] + \{x\}: 2[x]=[x]+2{x}2[x] = [x] + 2\{x\}, so [x]=2{x}[x] = 2\{x\}. The right side lies in [0,2)[0, 2), and the left side is an integer, so [x]∈{0,1}[x] \in \{0, 1\}. If [x]=0[x] = 0, then {x}=0\{x\} = 0 and x=0x = 0. If [x]=1[x] = 1, then {x}=12\{x\} = \frac{1}{2} and x=32x = \frac{3}{2}. Both satisfy the original equation (0=00 = 0, and 2=32+122 = \frac{3}{2} + \frac{1}{2}). The technique is to split xx into its integer and fractional parts, then use the fact that one is an integer and the other lies in [0,1)[0, 1) to reduce to a few cases. Brain-Teaser 3: How many real solutions does ∣x−1∣+∣x−3∣=2|x - 1| + |x - 3| = 2 have? What about ∣x−1∣+∣x−3∣=5|x - 1| + |x - 3| = 5 and ∣x−1∣+∣x−3∣=1|x - 1| + |x - 3| = 1? Solution: Read ∣x−1∣+∣x−3∣|x - 1| + |x - 3| as the total distance from xx to the points 11 and 33. For any xx between 11 and 33 that total is exactly the gap, 22, and outside that segment it is more than 22. So the equation with 22 holds for every x∈[1,3]x \in [1, 3], infinitely many solutions. With 55, there is one solution on each side: x=−12x = -\frac{1}{2} and x=92x = \frac{9}{2}. With 11, there are no solutions, because the total distance can never be less than 22. This is the triangle inequality in action: ∣x−1∣+∣3−x∣≥∣(x−1)+(3−x)∣=2|x - 1| + |3 - x| \ge |(x - 1) + (3 - x)| = 2, with equality exactly when x−1x - 1 and 3−x3 - x have the same sign, which is the interval [1,3][1, 3].

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Special Functions: Modulus, Greatest Integer, Fractional Part and Signum — DPP

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